Recent content by trow4

Marvelous, that indeed does the trick. However, if the polynomial is of a higher order (see below) I again run into trouble. For instance, if I want to find the real root of 0.1 + 0.5 (1-x)^4 - (1/y)=0, (which you can easily solve for x=1- [ 2 (1-0.1 y ) / y ]^(1/4) ), I again use sol =...

I am trying to solve 0.125 + 0.5 (1-x)^3 - (12.5/y)==0 for x, when y is real and y>0. I thus want to find x= 1- 0.63 ((100-y)/y)^(1/3), so that if y=100, x=1. Mathematica's Solve yields 3 roots: sol=Solve[0.125 + 0.5 (1 - x)^3 - (12.5/y) == 0, x] Root 1: 1.+((0.+0. I) y^(1/3))/(-100.+1...